I've been struggeling for quite a long time with problems like these:

Consider $V = \{ x \in \mathbb{R^5}: 2x_1 - x_4 < 7 \}$ Explain: Is $V$ convex? Is $V$ open? Is $V$ closed? Is $V$ bounded? Is $V$ compact?

I know for proving convexity that you can use the $\lambda x + (1-\lambda)y$-definition. But these steps take quite some time and I think you have to apply these steps if you want to prove whether a set is convex, not for explaining. So my question to you is: how do you explain that sets like the one above are convex, closed etc.?

The book and lecture notes that I use only provide some general definitions of convexity, closed, open, bounded, compact but does not offer clear examples that show how you can tackle questions like the one above. I have the feeling you should be able to see in a blink of the eye if such a set is convex, closed, open etc. But I just don't see it yet how and that bothers me.


1 Answer 1


The set $V$ is open as the preimage of the open set $(- \infty , 7)$ w.r.t. the continuous map $f(x) = 2x_1-x_4$. To prove convexity, you have to check that for any $\lambda \in (0,1)$ and $x,y \in V$, $\lambda x + (1-\lambda)y \in V$. I'll leave this as an exercise.

Also $V$ is clearly not compact, since it's open. It's also easy to show that it's not bounded.

  • $\begingroup$ Since $f$ is linear, your first argument even shows the convexity of $V$. $\endgroup$
    – gerw
    Oct 11, 2016 at 17:18
  • $\begingroup$ @gerw: Thanks for pointing that out! $\endgroup$ Oct 11, 2016 at 19:28
  • $\begingroup$ @gerw does that not contradict the property ''A linear combination $\sum_{k=1}^m \lambda_k x_k$ is called a convex combination if $\lambda_k \geq 0$" since $\lambda_4 = -1$? $\endgroup$
    – user377352
    Oct 12, 2016 at 10:16

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